Lecture 22
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Web Lecture 22
Class Lecture 18-Thursday 3/21/2013
Multiple Reactions with Heat Effects
2
Multiple Reactions with Heat Effects
PFR/PBR:
dT

dV
UaTa  T     rij H Rxij 
n
i 1
m
F C
j 1
j
Pj
CSTR:
UATa  T   FA0  CPi i T  T0  rij H Rxij T   0
3
m
q
j1
i 1
These equations are coupled with the mole
balances and rate law equations.
Multiple Reactions with Heat Effects
Multiple Reactions
Make sure it is in respect to A; Subscripts must agree
dT Q g  Q r

dV  Fi C Pi
4
Q g   rijH Rxij  r1A H Rx1A r2A H Rx 2A
Multiple Reactions with Heat Effects
Multiple Reactions
1) Mole Balances:– every species (no conversion!)
2) Rate Laws:
– relative rates
– net rates
3) Stoichiometry:
5
CA  CT 0
FA T0
y
FT T
dy   FT T

dW 2 y FT 0 T0
Multiple Reactions with Heat Effects
Multiple Reactions
4) Heat Effects:
dT Q g  Q r

dV  Fi C Pi
Q g  heat produced
Q r  heat removed
Q g   rij H Rxij
Q r  Ua(T - Ta )
6
(must havematchingi, j)
Multiple Reactions with Heat Effects
4) Heat Effects:
dT Q g  Q r

dV  Fi C Pi
Q g  r1A H R1A  r2 A H R 2 A
Q r  Ua T  Ta 
FC
i
Pi
 FA C PA  FB C PB  FC C PC  FD C PD
dTa Ua T  Ta 

 i C Pcool
dV
m
5) Parameters
7
E1 , E 2 , FA0 , Ua, ...etc
Multiple Reactions with Heat Effects
4) Heat Effects:
dT Qg  Qr

dV  Fi C Pi
23
H Rx1 A  12kJ /(mole of A reacted in reaction 1)
H Rx 2 B   8kJ /(mole of B reacted in reaction 2)
Qg  r1 A H Rx1 A  r2 B H Rx 2 B
Use relative rates of reaction
to get r2B in terms of the rate law that is given for reaction 2,
e.g. ,
8
-r
(2)
2A
k
3 A  2B - -  2D
3
2A
C C
A
B
then r2B
2
 r2A
3
The complex gas phase reactions
(1)
A  2B  C
r1A  k 1ACAC2B
H Rx1B  15,000 cal mol B
(2)
A  C  2D
r2C  k 2C CACC
H Rx 2A  10,000 cal mol A
take place in a 10 dm3 PFR with a heat exchanger. Plot the temperature, concentrations, molar
flow rates down the length of the reactor for the following operations. E.g., Note any maximums
minimums on your plot along with how they change for the different types of operations.
or
(a) Adiabatic operation
(b) Heat exchange with constant Ta
(c) Co current heat exchange
(d) Counter current heat exchange
(e) For parts (c) and (d), plot Qr and Qg down the length of the reactor. What do you observe?
Additional Information
C PA  10 cal mol K
C PC  30 cal mol K
C PB  10 cal mol K
C PD  20 cal mol K

 s K at 300K and E 1  8,000 cal mol
2
k 2C  2 dm 3 mol s K at 300K and E 2  12,000 cal mol
k 1A  40 dm 3 mol
C T 0  0.2 mol dm 3
9
2
, C PCool  1cal g K
ÝCool 20g s
, m
Ua  80 cal dm 3 s K , Tao  325K , T0  300K
FA0  5 mol s , FB0  10 mol s , FC0  0 , FD0  0
10
11
12
13
Multiple Reactions with Heat Effects
Multiple Reactions
4) Heat Effects:
UaT  Ta 
dT
  rA  H Rx  
dV
 FiCPi
dT Qg  Qr

dV  Fi C Pi
Qg   rij H Rij
14
Qg  heat produced
Qr  heat removed
(must have matching i, j)
Multiple Reactions with Heat Effects
in a PFR and CSTR
Examples:
(1) A  2B C
r1A  k1A CA CB2
and
H R1A  20,000 cal mol A

(2) 2A  3C D


r2C  k2C CA2 CC3
and
15
H R 2A  10,000 cal mol A
Example A: Liquid Phase CSTR
(1) A  2 B  C
r1A  k1A CA CB2
NOTE: The specific reaction rate k1A is defined with respect to
species A.

(2) 3C  2 A  D
r2C  k2C C C
3
C
2
A
NOTE: The specific reaction rate k2C is defined with respect to
species C.
16

Example A: Liquid Phase CSTR
The complex liquid phase reactions take place in a
2,500 dm3 CSTR. The feed is equal molar in A and B
with FA0=200 mol/min, the volumetric flow rate is 100
dm3/min and the reation volume is 50 dm3.
Find the concentrations of A, B, C and D existing in
the reactor along with the existing selectivity.
Plot FA, FB, FC, FD and SC/D as a function of V
17
Example A: Liquid Phase CSTR
Solution
Liquid Phase CSTR
1) Mole Balances:
(1)
f CA   0CA 0  0CA  rAV
(2)
f CB   0CB 0  0CB  rBV
(3)
f CC   0CC  rCV
(4)
f CD   0CD  rDV

18

2) Net Rates:
(5)
rA  r1A  r2A

Example A: Liquid Phase CSTR
3) Stoichiometry:
(16)
CA  FA 0
(17)
CB  FB 0
(18)
CC  FC 0
(19)
CD  FD 0
4) Parameters:
19
(20)
0  100dm3 min
(21)
k1A  10 dm mol min
(22)
3

 15 dm
3
k 2C

mol 
2
4
min
Example B: Liquid Phase PFR
Takes place in a PFR. The feed is equal molar in A and
B and FA0=200 mol/min and the volumetric flow rate is
100 dm3/min. The reaction volume is 50 dm3 and the
rate constants are:
k1A  10 dm mol  min
3
2
k2C  15 dm mol  min
3
4
Rate laws are the same as Example A.

Plot FA, FB, FC, FD and SC/D as a function of V.
20
Example B: Liquid Phase PFR
1) Mole Balances:
21
Example B: Liquid Phase PFR
2) Net Rates:
(5)
(6)
(7)
(8)
rA  r1A  r2A
rB  r1B
rC  r1C  r2C
rD  r2D
2) Rate Laws:

2
r1A  k1A CA CB
(9)
22
(10)
r2C  k2C C C
2
A
3
C
Example B: Liquid Phase PFR
2) Relative Rates:
r1A r1B r1C


1  2 1
(11)
(12)
r1B  2r1A
r1C  r1A
r2A r2C r2D


2 3 1
23
(13)
r2A  2 3r2C
(14)
r2D  1 3r2C

Reaction 1
Reaction 2
Example B: Liquid Phase PFR
2) Rate Laws:
r1 A   k1 AC AC B2
r2C   k 2C C A2 CC3
rA  r1 A  r2 B
rC  r1C  r2C
r1C   r1 A
r2 A  2 / 3r2C
24
5 k1 A  k1 A1 expE1 R 1 T1  1 T  6
7  k 2C  k 2C 2 expE2 R 1 T2  1 T  8
9 rB  r1B 10
11 rD  r2 D 12
13 r1B  2r1 A 13
15 r2 D  1 / 3r2C 16
Example B: Liquid Phase PFR
3) Stoichiometry:
C A  CT 0
FA T0
y
FT T
17
CC  CT 0
FC T0
y
FT T
19
FT  FA  FB  FC  FD
25
21
C B  CT 0
FB T0
y
FT T
18
C D  CT 0
FD T0
y
FT T
20
dy   FT T

dV
2 y FT 0 T0
22
Multiple Reactions with Heat Effects
4) Heat Effects:
Qg  Qr
dT

dV  Fi C Pi
dTa UaT  Ta 

 i C Pcool
dV
m
23
26
24
25
Qr  UaT  Ta 
 CP  FACPA  FBCPB  FC CPC  FDCPD 27 
Qg  r1 A H R1 A  r2 A H R 2 A
Parameters:
26
E1 , E2 , FA0 ,....
Selectivity
If one were to write SC/D=FC/FD in the Polymath
program, Polymath would not execute because at V=0,
FC=0 resulting in an undefined volume (infinity) at V=0.
To get around this problem we start the calculation 10-4
dm3 from the reactor entrance where FD will not be zero
and use the following IF statement.
(15)
27

S˜C D
FC 
 if V  0.001 then   else 0
FD 
Selectivity
3) Stoichiometry:
(16)
CA  FA 0
(17)
CB  FB 0
(18)
CC  FC 0
(19)
CD  FD 0
Parameters:
28
(20)
0  100dm3 min
(21)
k1A  10 dm3 mol min
(22)
k 2C

 15 dm

mol 
2
3
4
min
End of Web Lecture 22
Class Lecture 18
29